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Z score to p values

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lou
lou el 7 de Dic. de 2016
Comentada: Germano Gallicchio el 28 de Feb. de 2024
I have a large matrix 1xN containing z values. I would like to know how to turn these z scores to p values using normcdf function?
How to obtain p values both for one-tailed and two-tailed p values using normcdf?
Many thanks in advance!

Respuesta aceptada

Star Strider
Star Strider el 7 de Dic. de 2016
If I remember correctly, the probability of a one-tailed test is twice the probability of a two-tailed test, so:
p_one = 2*normcdf(z_vector);
p_two = normcdf(z_vector);
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Star Strider
Star Strider el 30 de Mzo. de 2017
It depends upon the hypothesis you are testing.
Noah
Noah el 23 de Jul. de 2021
Note that the accepted answer is backwards, unless you mean something strange by your hypothesis. The probability of one-tailed test is HALF the probability of a two-tailed test. The area under a bell curve on one side is half the area on both sides.
p_oneTailed = normcdf(z_vector);
p_twoTailed = 2*normcdf(z_vector);

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Más respuestas (1)

Ziwei Liu
Ziwei Liu el 18 de Ag. de 2023
Two-tailed p value should actually be 2 * (1 - normcdf(z)).
normcdf(z) gives the area under curve on the left side of z. This is not p value. One-tailed p value should be the area on the right side, which is (1 - normcdf(z)).Two-tailed p value should be the double of that.
You can use the arrayfun function to compute p value for each entry in your z score matrix. i.e. p = arrayfun(@(x) 2*(1-normcdf(x)), ZScoreMatrix).
  1 comentario
Germano Gallicchio
Germano Gallicchio el 28 de Feb. de 2024
For this to work with negative z scores, you also need to take the absolute value of z:
z = [-2.58 -1.96 -1.65 0 1.65 1.96 2.58]; % vector of z scores
p = 2 * (1 - normcdf(abs(z))); % vector of associated pvalues
disp([z' p'])
-2.5800 0.0099 -1.9600 0.0500 -1.6500 0.0989 0 1.0000 1.6500 0.0989 1.9600 0.0500 2.5800 0.0099

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